# The Ultimate Newcomb’s Problem

You see two boxes and you can ei­ther take both boxes, or take only box B. Box A is trans­par­ent and con­tains \$1000. Box B con­tains a visi­ble num­ber, say 1033. The Bank of Omega, which op­er­ates by very clear and trans­par­ent mechanisms, will pay you \$1M if this num­ber is prime, and \$0 if it is com­pos­ite. Omega is known to se­lect prime num­bers for Box B when­ever Omega pre­dicts that you will take only Box B; and con­versely se­lect com­pos­ite num­bers if Omega pre­dicts that you will take both boxes. Omega has pre­vi­ously pre­dicted cor­rectly in 99.9% of cases.

Separately, the Numer­i­cal Lot­tery has ran­domly se­lected 1033 and is dis­play­ing this num­ber on a screen nearby. The Lot­tery Bank, like­wise op­er­at­ing by a clear known mechanism, will pay you \$2 mil­lion if it has se­lected a com­pos­ite num­ber, and oth­er­wise pay you \$0. (This event will take place re­gard­less of whether you take only B or both boxes, and both the Bank of Omega and the Lot­tery Bank will carry out their pay­ment pro­cesses—you don’t have to choose one game or the other.)

You pre­vi­ously played the game with Omega and the Numer­i­cal Lot­tery a few thou­sand times be­fore you ran across this case where Omega’s num­ber and the Lot­tery num­ber were the same, so this event is not sus­pi­cious.

Omega also knew the Lot­tery num­ber be­fore you saw it, and while mak­ing its pre­dic­tion, and Omega like­wise pre­dicts cor­rectly in 99.9% of the cases where the Lot­tery num­ber hap­pens to match Omega’s num­ber. (Omega’s num­ber is cho­sen in­de­pen­dently of the lot­tery num­ber, how­ever.)

You have two min­utes to make a de­ci­sion, you don’t have a calcu­la­tor, and if you try to fac­tor the num­ber you will be run over by the trol­ley from the Ul­ti­mate Trol­ley Prob­lem.

Do you take only box B, or both boxes?

• As wed­ifrid said, this is ap­prox­i­mately trans­par­ent New­comb plus dis­trac­tions. Given Eliezer’s clar­ifi­ca­tions (Omega knows the lot­tery num­bers and is ac­cu­rate even when the Omega and lot­tery num­bers match), I’ll ask how two al­gorithms would perform against Omega: OneBoxBot, which always one-boxes, and Con­di­tion­alBot, which one-boxes un­less the lot­tery and Omega num­bers match, in which case it two-boxes. I’ll ig­nore the tiny er­ror rates in com­put­ing pay­offs.

Case 1: the lot­tery is go­ing to out­put a prime num­ber.

Against OneBoxBot, Omega de­liv­ers a prime num­ber, which may or may not match the lot­tery num­ber. OneBoxBot gets a pay­off of \$1MM.

Against Con­di­tion­alBot, Omega must de­liver a prime num­ber differ­ent from the lot­tery num­ber (if it matched the lot­tery Omega would be hand­ing over a prime num­ber with a pre­dicted re­sponse of two-box­ing from Con­di­tional). So Con­di­tion­alBot one-boxes and gets a pay­off of \$1MM.

Case 2: the lot­tery will out­put a com­pos­ite num­ber.

OneBoxBot will one-box, so Omega must provide it with a prime num­ber (which will not match the lot­tery). So OneBoxBot gets a pay­off of \$3MM ev­ery time the lot­tery ran­domly out­puts a com­pos­ite num­ber.

Against Con­di­tion­alBot, Omega has two choices. It could de­liver a prime num­ber which does not match the lot­tery. This would lead Con­di­tion­alBot to one-box and get a to­tal pay­off of \$3MM, the same as OneBoxBot. Alter­na­tively, Omega could match the com­pos­ite lot­tery num­ber, lead­ing Con­di­tion­alBot to two-box and get a pay­off of \$2,001,000, \$999,000 worse than OneBoxBot’s pay­off.

So: both al­gorithms re­ceive lot­tery wins with equal fre­quency, OneBoxBot always performs at least as well as Con­di­tion­alBot across the pos­si­bil­ities, and Con­di­tion­alBot performs worse when the lot­tery out­puts a com­pos­ite and Omega chooses to match it. Thus we should not adopt Con­di­tion­alBot over OneBoxBot and should one-box faced with the Ul­ti­mate New­comb prob­lem.

• There is a big and im­plicit step that is worth ex­pli­cat­ing here, be­cause most peo­ple who first ap­proach New­comb-like prob­lems miss it com­pletely:

TREAT HUMANS AS BOTS

By a bot I mean an al­gorithm, of course. An al­gorithm Omega can an­a­lyze for all pos­si­ble com­bi­na­tion of in­puts.

That this step is valid fol­lows from the prob­lem’s stipu­la­tion that Omega can pre­dict your ac­tions. In other words, it knows your out­put for any com­bi­na­tion of in­puts it cares to give you.

Whether Omega does this by run­ning your al­gorithm in a sand­box or by an­a­lyz­ing your code does not af­fect the an­swer to the puz­zle, since the end re­sult is the same. But the sand­box­ing ver­sion can of­ten make it eas­ier to find the solu­tion, be­cause it lets one rely on the Reflec­tive Equil­ibrium of sorts: you can­not tell when de­cid­ing what to do whether you are in an Omega’s simu­la­tion of you or not, so you may as well as­sume that you are.

TL;DR: to Omega, you are a bot, so write down all rele­vant al­gorithms and an­a­lyze/​run them be­fore pick­ing a win­ning one.

• This is what I did too. One big ad­van­tage is that it changes Omega’s pre­dic­tive abil­ities from mys­te­ri­ous magic to a sim­ple pro­cess that it’s pos­si­ble to com­pletely analyse with­out get­ting con­fused.

• I’m afraid that, as writ­ten, I can­not an­swer the prob­lem’s fi­nal ques­tion, as by the time it was asked...

… I’d been hit by the trol­ley.

• Yeah, same. But I figured it wouldn’t vi­o­late the spirit of the ex­er­cise to pre­tend to turn the clock back two min­utes and only set it go­ing again once I un­der­stood the whole post!

(I then spent about 90 sec­onds men­tally flailing about and ty­ing my­self in knots, be­fore notic­ing I only had a few sec­onds left. At which point I told my­self to find the sim­plest, dumb­est solu­tion that might pos­si­bly work; bet­ter to give a sub­op­ti­mal an­swer than to get splat­tered by a trol­ley. And the sim­plest, dumb­est solu­tion is that the Lot­tery num­ber & pay­out are wholly in­de­pen­dent of what Omega or I do, so I can com­pletely ig­nore it to re­duce the prob­lem to the usual New­comb’s prob­lem. Hence take only box B.)

• Has any­one ac­tu­ally tried to an­swer the Ul­ti­mate Trol­ley Prob­lem? I’m think­ing right.

• I trust that ob­serv­ing that it’s not a mul­ti­ple of 2, 3, 5, or 11 doesn’t count.

• Depend­ing on what does or doesn’t count, it might be pos­si­ble to be sure of whether 1033 is prime with­out “try­ing to fac­tor” it (i.e., by check­ing whether it’s di­visi­ble (with­out ac­tu­ally carry out the di­vi­sions) by any prime num­ber less than sqrt(1033)).

So I in­ter­preted “try to fac­tor it” pes­simisti­cally as “try to figure out for sure whether it’s prime”.

• Right. And, in­deed, when I need to check in my head whether a small­ish num­ber like 1033 is prime, most of what I do isn’t fac­tor­ing. E.g.: 1033+17 = 1050 = 50.21, rul­ing out {2,3,5,7,17} in a sin­gle go but not iden­ti­fy­ing any fac­tors or (even ap­prox­i­mate) quo­tients. 1033-13 = 1020 = 20.51, rul­ing out {2,3,5,13,17} in­stead. Etc.

I also as­sumed that Eliezer’s in­ten­tion was that you shouldn’t be do­ing this sort of thing.

[EDITED to fix for­mat­ting screwage—as­ter­isks treated as markup rather than mul­ti­pli­ca­tion—my apolo­gies for any mys­tifi­ca­tion this caused.]

• I’m cu­ri­ous, what’s the method you’re us­ing there?

• (Ini­tial re­mark: damn, I see that what I wrote got screwed up by as­ter­isks get­ting treated as markup char­ac­ters; per­haps ev­ery­thing would have been clear with­out that. Will fix once I’ve finished writ­ing this.)

I’m not sure it re­ally war­rants dig­nify­ing with the term “method”, but:

The re­la­tions I’m look­ing for are of the form n=a+b or n=a-b where a,b (1) are easy to fac­tor be­cause they’re prod­ucts of small primes, and (2) have no com­mon fac­tor. In that case, you know that n isn’t a mul­ti­ple of any of those prime num­bers—so, e.g., if a, but not b, is a mul­ti­ple of p, then a+b and a-b are not mul­ti­ples of p.

The eas­iest case is where b is it­self a small prime num­ber like 17 or 13. Why 17 and 13? Be­cause the num­ber ends in 3, so sub­tract­ing some­thing end­ing in 3 or adding some­thing end­ing in 7 will give us (an easy-to-deal-with fac­tor of 10, and) some­thing nicely smaller. In some cases it’s eas­ier to re­move digits from the start of the num­ber rather than the end; for a triv­ial ex­am­ple, 1033 isn’t a mul­ti­ple of 103 be­cause it’s 1030 + 3.

Let’s do 1033 a bit more thor­oughly. It’s be­tween 32^2=1024 and 33^2=1089, so we need to check primes up to 31. 1033=1050-17 rules out 2,3,5,7,17. 1033=1020+13 rules out 2,3,5,13,17. 1033=1000+33 rules out 2,3,5,11. (That’s all the primes up to 17 done, leav­ing 19,23,29,31.) 1033=1010+23 rules out 2,5,(101),23. At this point, ac­tu­ally, I’d try di­vid­ing by 19,29,31, but let’s carry on. For 31 we’ll sub­tract 93: 1033=940+93, and 940=2.2.5.47 so we’ve ruled out 31. For 19 we’ll sub­tract 950, leav­ing 83 which is prime, so 19 is ruled out. For 29 I sup­pose the best we can do is to sub­tract 870, leav­ing 163 which I hap­pen to know is prime be­cause of exp(pi sqrt(163)) but in any case clearly isn’t a mul­ti­ple of 29. So we’re done and 1033 is prime. And I didn’t fac­tor­ize 1033 un­less prov­ing some­thing is prime by a round­about route counts as fac­tor­iz­ing it. If one of these things had turned out to have the same prime num­ber di­vid­ing both sum­mands, there’d still have been ex­tra work to do to get the quo­tient.

• Primes less than sqrt(1033) for which I know of no re­ally ob­vi­ous tricks (i.e. the digit adding tricks for them aren’t so sim­ple one can triv­ially do them in your head): 7, 13, 17, 19, 23, 29, 31

Also, since the sce­nario said we’ve done this many times and we haven’t been trol­leyed yet, it can’t be all that easy to get trol­leyed.

(eta: why the −1? Both points seem solid to me, even in the light of the ad­di­tional trick be­low—there are sev­eral pos­si­ble fac­tors re­main­ing, and it’s not like I was enu­mer­at­ing these dur­ing my 2 min­utes. More­over, the 1001 trick works for 1033, but doesn’t help so much with other typ­i­cal num­bers of that gen­eral mag­ni­tude—say, 1537, that it’s some­thing you’re li­able to do by ac­ci­dent)

• Also, since the sce­nario said we’ve done this many times and we haven’t been trol­leyed yet, it can’t be all that easy to get trol­leyed.

Maybe you get run over by the trol­ley only if you try to fac­tor Omega’s num­ber, and all the times be­fore this Omega’s num­ber and the lot­tery num­ber were differ­ent, and there’s no much point in try­ing to fac­tor the lot­tery num­ber in that case since (as­sum­ing lin­ear util­ity of money) it has no rele­vance on how many boxes you should take.

• 1001 = 7 x 11 x 13, so each of those di­vides 1033 iff it di­vides 32 (and di­vides 314,159,265 iff it di­vides 265 − 159 + 314).

• Omega also knew the Lot­tery num­ber be­fore you saw it, and while mak­ing its pre­dic­tion, and Omega like­wise pre­dicts cor­rectly in 99.9% of the cases where the Lot­tery num­ber hap­pens to match Omega’s num­ber.

(Omega’s num­ber is cho­sen in­de­pen­dently of the lot­tery num­ber, how­ever.)

I’m not sure that both these state­ments can be true at the same time.

What does Omega pick if my al­gorithm is, “if the num­ber ends in 3, two-box, else one-box”? Seems it will be right if it ei­ther picks a com­pos­ite end­ing in 3 (where I don’t get the mil­lion) or a prime not end­ing in 3 (where I do), so how does it de­cide?

• Maybe I’m con­fused, but can’t it just… pick one how­ever it wants? As long as the fi­nal tally comes to “cor­rect in 99.9% of cases” the prob­lem is satis­fied and it doesn’t mat­ter how it de­cided, does it? Why does it mat­ter if there are muli­ple cor­rect choices?

Edit: Nev­er­mind, I see what you mean. If your al­gorithm de­pends on the Lotto, and Omega de­pends on your al­gorithm, then Omega de­pends on the Lotto. You can ac­tu­ally in­fluence how of­ten this “same num­ber” sce­nario hap­pens via your al­gorithm.

• Ex­cel­lent point.

• A plau­si­ble setup:

There are two equal-sized groups of large num­bers:

• Prime numbers

• Com­pos­ite num­bers with­out any easy factorization

… such that you can’t be ex­pected to dis­t­in­guish them in the al­lowed time.

The lot­tery works by pick­ing at one of the num­bers at ran­dom.

Omega’s al­gorithm: for each num­ber in those groups, pre­dict whether you would two-box (given the already-de­ter­mined lot­tery num­ber). In the set of prime num­bers for which you one-box, and com­pos­ite num­bers for which you two-box, pick a num­ber at ran­dom and show it to you (so if it pre­dicts you always two-box, it’s sure to pick a com­pos­ite num­ber, etc.).

In that setup, if your al­gorithm is “if the num­ber ends in 3, two-box, else one-box”, then Omega will just pick a ran­dom num­ber among the com­pos­ites end­ing in 3 and the primes end­ing in 1, 3, 7, or 9.

And if your al­gorithm is “If the num­ber is the same as the lot­tery, two-box, else, one-box”, then yeah the cho­sen num­ber will not be com­pletely in­de­pen­dent of the lot­tery num­ber (Omega can only pick the lot­tery num­ber if it’s com­pos­ite, or if it makes a pre­dic­tion er­ror), but that looks in­de­pen­dent enough for the spirit of the post.

• I’m not sure that both these state­ments can be true at the same time.

If you take the sec­ond state­ment to mean, “There ex­ists an al­gorithm for Omega satis­fy­ing the prob­a­bil­ities for cor­rect­ness in all cases, and which some­times out­puts the same num­ber as NL, which does not take NL’s num­ber as an in­put, for any al­gorithm Player tak­ing NL’s and Omega’s num­bers as in­put,” then this …seems… true.

I haven’t yet seen a com­ment that proves it, how­ever. In your ex­am­ple, let’s as­sume that we have some al­gorithm for NL with some speci­fied prob­a­bil­ity of out­putting a prime num­ber, and some speci­fied prob­a­bil­ity it will end in 3, and maybe some dis­tri­bu­tion over mag­ni­tude. Then Omega need only have an al­gorithm that out­puts com­bi­na­tions of prime­ness and 3-end­ed­ness such that the prob­a­bil­ities of out­comes are satis­fied, and which some­times pro­duces co­in­ci­dences.

For some al­gorithms of NL, this is clearly im­pos­si­ble (e.g. NL always out­puts prime, c.f. a Player who always two-boxes). What seems less cer­tain is whether there ex­ists an NL for which Omega can always gen­er­ate an al­gorithm (satis­fy­ing both 99.9% prob­a­bil­ities) for any al­gorithm of the Player.

This is to say, what we might have in the state­ment of the prob­lem is ev­i­dence for what sort of al­gorithm the Nat­u­ral Lot­tery runs.

Per­haps what Eliezer means is that the prime­ness of Omega’s num­ber may be in­fluenced by the prime­ness NL’s num­ber, but not by which num­ber speci­fi­cally? Maybe the sec­ond state­ment is meant to sug­gest some­thing about the like­li­hood of there be­ing a co­in­ci­dence?

• CDT:

Two box, ob­vi­ously.

EDT:

As­sum­ing this is your last game, two box. Two box­ing is ev­i­dence that 1033 is com­pos­ite, so you’ll get more money.

If you will con­tinue play­ing for a long time, one box. This is ev­i­dence that you will go with the “always one box” strat­egy, which will re­sult in more money. More gen­er­ally, it is ev­i­dence that you will go with a TDT-style strat­egy more of­ten in the fu­ture, and get higher pay­outs as a re­sult.

TDT:

One box. The always one box strat­egy has the high­est pay­out.

I’m not sure if I have the right ter­minol­ogy with TDT, but these are the three ob­vi­ous moves and the rea­son­ing for them.

• EDT:

As­sum­ing this is your last game, two box.

Ahh, good point. This ex­plains the (likely) mo­ti­va­tion for Eliezer to con­trive this sce­nario. It’s a case where one box­ing is the right choice but even EDT gets it wrong. Usu­ally at least one of CDT or EDT gets it right.

• One-box? I would have said two-box, un­der the bizarre the­ory that I can thereby cause the num­ber to be com­pos­ite.

Checks… hmm. Well, that was un­likely.

• I ini­tially thought two-box, but on think­ing about it more, I’m go­ing for one-box.

For sim­ple num­bers, let’s sup­pose that the lot­tery has a 50% chance of choos­ing a prime num­ber, and that if Omega could se­lect the same num­ber as the lot­tery, he’ll do so with 10% prob­a­bil­ity.

Three sim­ple strate­gies:

1) Always one-box: Gets Omega’s pay­out ev­ery time, wins the lot­tery 50% of the time. Aver­age to­tal pay­out \$2M. (num­bers are the same 10% of the time when the lot­tery is ‘prime’)

2) Always two-box: Omega never pays out, wins the lot­tery 50% of the time. Aver­age to­tal pay­out \$1.001M. (num­bers are the same 10% of the time when the lot­tery is ‘com­pos­ite’)

3) Nor­mally one-box, two-box when num­bers are the same. Omega pays out 95% of the time. Lot­tery pays out 50% of the time. Aver­age to­tal pay­out \$1.95M. (Num­bers are the same 10% of the time when the lot­tery is ‘com­pos­ite’)

The trick is that the ques­tion tries to lead you to the wrong coun­ter­fac­tual by draw­ing your at­ten­tion to the situ­a­tion where the num­bers are the same. Whether you see the num­bers be­ing the same de­pends on your de­ci­sion. In the coun­ter­fac­tual world where you de­cide some­thing else, the lot­tery num­ber doesn’t change to match Omega’s pre­dic­tion. In­stead, in the coun­ter­fac­tual world, the lot­tery num­ber and Omega’s num­ber are differ­ent.

• I don’t see the rele­vance. The com­menter con­tem­plated a hy­po­thet­i­cal sce­nario through ab­stract think­ing, there’s no em­piri­cism here.

• Ac­tu­ally do­ing the math, rather than just rely­ing on in­tu­ition about what sounds right.

• I’m not sure where you got that 95% num­ber from for your strat­egy #3; it sounds like the “both num­bers are the same” situ­a­tion only hap­pens once ever sev­eral thou­sand of runs.

Any­way, if you’re us­ing strat­egy 1, then if the two num­bers are the same, that means that the num­ber is prime, and your pay­out for this sce­nario is only 1 mil­lion dol­lars (you lost the lot­tery). If you’re us­ing strat­egy 3, then that means the num­ber is not prime, and the pay­out is \$2.001 mil­lion dol­lars (the num­ber is not prime, be­cause you’re go­ing to dou­ble box.)

There is no differ­ence be­tween strat­egy 1 and strat­egy 3 ex­cept in the one sce­nario where both num­bers are the same, and the one sce­nario where both num­bers are the same, strat­egy 3 is bet­ter. There­fore, strat­egy 3 is always bet­ter.

• Although the ti­tle was origi­nally se­lected in jest, I think this may ac­tu­ally be the Ul­ti­mate New­comb’s Prob­lem be­cause it tempts the largest num­ber of peo­ple—EDTers, CDTers, and ap­par­ently a sub­stan­tial por­tion of LWers—to two-box.

• I was sur­prised that so many peo­ple wanted to two-box. Maybe they thought you were only a level 1 agent? I mean, I ad­mit that was part of my de­liber­a­tion, but the math backs me up, so....

• If Omega main­tains a 99.9% ac­cu­racy rate against a strat­egy that changes its de­ci­sion based on the lot­tery num­bers, it means that Omega can pre­dict the lot­tery num­bers. There­fore, if the lot­tery num­ber is com­pos­ite, Omega has mul­ti­ple choices against an agent that one-boxes when the num­bers are differ­ent and two-boxes when the num­bers are the same: it can pick the same com­pos­ite num­ber as the lot­tery, in which case the agent will two-box and earn 2,001,000, or it can pick a differ­ent prime num­ber, and have the agent one-box and earn 3,001,000. It seems like the agent that one-boxes all the time does bet­ter by elimi­nat­ing the cases where Omega se­lects the same num­ber as the lot­tery, so I would one box.

• The op­ti­mal choice if you’re just told you’re gonna play this game a bunch of times is one-box­ing. (Edit: this is be­cause be­cause try­ing to con­trol the lot­tery num­ber will just re­sult in less games where the num­bers are the same). How­ever, if you’re guaran­teed a run where the num­bers come up the same, then there’s some hid­den con­trol, via the thought ex­per­i­ment it­self. If I two box, Omega picked a com­pos­ite, so in or­der for the sce­nario to have oc­curred at all (which by the thought ex­per­i­ment it’s guaran­teed to have) the lot­tery must have come up with a com­pos­ite. Cal­ling some­thing ran­dom doesn’t make it ran­dom. I prob­a­bly wouldn’t have no­ticed if I hadn’t re­cently read GAZP vs GLUT.

• The Ul­ti­mate New­comb’s Problem

This seems to be just trans­par­ent New­comb’s prob­lem com­bined with some re­dun­dant words. Un­less I’ve missed some­thing in the fine print this is a sim­ple “one box” situ­a­tion.

• All de­ci­sion the­ory prob­lems are sim­ple ones com­bined with some con­fu­sion that a more en­light­ened mind would see as re­dun­dant and iso­mor­phic to the sim­ple.

• Not always. Some of them are just ill-defined or im­pos­si­ble, com­bined with some con­fu­sion to stop peo­ple notic­ing. (eg, a ver­sion of Trans­par­ent New­comb that doesn’t spec­ify what Omega does if a player de­cides to op­pose Omega’s pre­dic­tion)

• Then again, a more en­light­ened mind would know whether 1033 is prime.

• All de­ci­sion the­ory prob­lems are sim­ple ones com­bined with some con­fu­sion that a more en­light­ened mind would see as re­dun­dant and iso­mor­phic to the sim­ple.

I came to ap­pre­ci­ate this par­tic­u­lar ex­am­ple when it was ob­served that EDT and CDT both two box (de­spite this be­ing a bad de­ci­sion). Usu­ally one or the other gets it right and peo­ple at times take their pick from them on a prob­lem by prob­lem ba­sis (some even try to for­mal­ise this ‘take your pick’ al­gorithm). This sce­nario is in the bal­l­park of the sim­plest prob­lem that re­ally sets the de­ci­sion the­o­ries apart.

• I don’t un­der­stand why two-box­ing is a bad de­ci­sion when 99.9% of two-box­ers will have 2 mil­lion dol­lars and 99.9% of the one-box­ers will have only 1 mil­lion.

• I don’t un­der­stand why two-box­ing is a bad de­ci­sion when 99.9% of two-box­ers will have 2 mil­lion dol­lars and 99.9% of the one-box­ers will have only 1 mil­lion.

It is a mat­ter of de­ter­min­ing what is con­trol­led by the de­ci­sion pro­ce­dure of the player or Omega (who is in­fluenced by the player) and what is con­trol­led by ar­ith­metic and the whims of the per­son choos­ing which hy­po­thet­i­cal prob­lem to talk about. In this case win­ning or los­ing the lot­tery is pure luck while win­ning or los­ing with Omega’s game is de­ter­mined by the player’s de­ci­sion. While two-box­ing can change the ev­i­dence of whether you won the lot­tery it never in­fluences the lot­tery out­come one way or the other. On the other hand Omega’s choice is di­rectly de­ter­mined by the player’s de­ci­sion pro­ce­dure.

It may be helpful to con­sider an­other hy­po­thet­i­cal game which has similar difficul­ties and which I had pre­vi­ously been us­ing for my pur­poses in the role of ‘Ul­ti­mate New­comb’s Prob­lem’. Con­sider:

• Take New­comb’s Problem

• Add ‘Trans­par­ent Boxes’ mod­ifi­ca­tion. (This step is op­tional—the pro­hi­bi­tion on fac­tor­ing makes the pro­cess some­what opaque.)

• In­clude ran­dom com­po­nent. Ac­cord­ing to some source con­sid­ered ran­dom Omega fills the big box as usual 99.99% of the time but the re­main­ing 0.01% of the time he in­verts his pro­ce­dure.

• Posit the prob­lem where the player finds him­self star­ing at an empty big box and a small box with \$1,000. Ask him whether or not he takes the small box.

Many peo­ple ad­vo­cate two-box­ing in such a situ­a­tion. I one box. I would ex­pect peo­ple who two box in Ran­dom­ized Trans­par­ent New­comb’s to also two box here for con­sis­tency. I would con­sider it odd for a RTN two-boxer to One box on Ul­ti­mate New­comb’s. In the above sce­nario I lose. I get \$1,000. But I lose \$1,000 1 time out of 10,000 and win \$1,000,000 the other 9,999 times. Some­one who two boxes po­ten­tially wins \$1,000 up to 9,999 times out of 10,0000 (find­ing the equil­ibrium re­sult there gets weird and de­pends on de­tails in the speci­fi­ca­tion).

Now I may talk of win­ning 9,999 times out of 10,000 but that can sound hol­low when the fact re­mains that I lose in the only ex­am­ple brought up. Why do I con­sider it ok to not win in this case? What makes me think it is ok to op­ti­mise for differ­ent situ­a­tions to the one that ac­tu­ally hap­pens? Depend­ing on the point of view this is be­cause I don’t at­tempt to con­trol ran­dom or I don’t at­tempt to con­trol nar­ra­tive causal­ity. Omega’s be­hav­ior I can in­fluence. I can­not in­fluence as­sumed ran­dom sources and if I fol­low the im­prob­a­bil­ity to the au­thor’s choice of hy­po­thet­i­cal then I choose not to op­ti­mize for sce­nar­ios based on how likely they are to be dis­cussed. It would be a differ­ent ques­tion if ev­i­dence sug­gested I was liv­ing in a physics based on nar­ra­tive causal­ity where one-in-a-mil­lion chances oc­cur nine times out of ten.

So in short, if the lot­tery doesn’t give me \$2M that is be­cause I am un­lucky but if Omega doesn’t give me \$1M it is be­cause I am a dum­b­ass. The differ­ence be­tween the two is sub­tle but crit­i­cally im­por­tant.

• Sorry for the de­lay to re­spond, I’ve been busy last cou­ple weeks.

I think you’ve con­vinced me re­gard­ing this. To dis­cuss my own per­spec­tive on this, in the past it took me quite a while be­fore I ‘got’ why one-box­ing is the right de­ci­sion in Trans­par­ent New­comb—it was only once I started think­ing of de­ci­sions as in­stances of a de­ci­sion the­ory/​de­ci­sion pro­ce­dure that I re­al­ized how a “los­ing” de­ci­sion may ac­tu­ally be part of what’s a “win­ning” de­ci­sion the­ory over­all—and that there­fore one-box­ing is the cor­rect strat­egy in Trans­par­ent New­comb.

I guess that in Ul­ti­mate New­comb, one-box­ing re­mains a win­ning de­ci­sion the­ory, though again the win­ning de­ci­sion the­ory is rep­re­sented in a seem­ingly ‘los­ing’ de­ci­sion. That I failed to get the cor­rect an­swer here means that though I had un­der­stood, I had not re­ally grokked the logic be­hind this—I be­haved too much as if EDT was cor­rect in­stead.

Thanks for guid­ing me through this. Much ap­pre­ci­ated!

• Cf. a bunch of com­ments in this thread ap­par­ently think­ing you can con­trol ar­ith­metic...

• It’s trans­par­ent New­comb’s prob­lem but of the “per­verse” va­ri­ety where even when you see that the mil­lion is miss­ing you’re sup­posed to not take the thou­sand. The typ­i­cal (e.g. in Good and Real) ver­sion just re­quires you to one-box when the mil­lion is there.

• It’s trans­par­ent New­comb’s prob­lem but of the “per­verse” va­ri­ety where even when you see that the mil­lion is miss­ing you’re sup­posed to not take the thou­sand. The typ­i­cal (e.g. in Good and Real) ver­sion just re­quires you to one-box when the mil­lion is there.

I do pre­fer the “per­verse” var­i­ant. The typ­i­cal ver­sion seems so triv­ial. I’ve already got defined-by-the-prob­lem cer­tainty about the pay­offs. A few pho­tons in my eyes adds lit­tle. To give the ‘per­verse’ var­i­ant more em­pha­sis I also like to add a small amount of ran­dom noise to Omega’s choice so that the “one box when box empty” sce­nario isn’t self-pre­vent­ing.

• Read­ing this was like hear­ing Vin­cent Price say­ing “your pay­off ma­tri­ces are use­less here! Aha­ha­haha!” That was a le­gi­t­i­mate source of epistemic dread.

It took me about n minute of ghea­vat it bire to fully grok the struc­ture of the prob­lem, ng which point V set­tled on two-box­ing, which in this case, with this num­ber, leaves me hold­ing a mere thou­sand dol­lars, be­cause this is one of the 0.01% of uni­verses where Omega was wrong.

I am pretty solidly sold on two-box­ing at the mo­ment, un­der the ab­surd-sound­ing premise that I can in­fluence whether or not the num­ber is prime with my de­ci­sion. I re­ally hope time travel isn’t pos­si­ble.

• Yes, I saw at­tempt­ing to am­bi­ently con­trol ba­sic ar­ith­metic as the ob­vi­ous solu­tion here, too.

If I one-box then (ig­nor­ing through­out the tiny prob­a­bil­ities of Omega be­ing wrong) the num­ber is prime. I re­ceive \$1M from Omega and \$0 from the Lot­tery.

If I two-box then the num­ber is com­pos­ite, Omega pays me \$1K, and the Lot­tery pays me \$2M.

There­fore I two-box.

• I think this line of rea­son­ing re­lies on the Num­ber Lot­tery’s choice of num­ber be­ing con­di­tional on Omega’s eval­u­a­tion of you as a one-boxer or two-boxer. The prob­lem de­scrip­tion (at the time of this writ­ing) states that the Num­ber Lot­tery’s num­ber is ran­domly cho­sen, so it seems like more of a dis­trac­tion than some­thing you should try to ma­nipu­late for a bet­ter pay­off.

Edit: Dis­trac­tion is definitely the wrong word. As ShardPhoenix in­di­cated, you might be able to get a bet­ter pay­off by mak­ing your one-box /​ two-box de­ci­sion de­pend on the out­come of the Num­ber Lot­tery.

• eval­u­a­tion of you as a one-boxer or two-boxer

Those are not the only pre­com­mit­ments one can make for this type of situ­a­tion.

• Of course! I meant to say that Richard’s line of thought was mis­taken be­cause it didn’t take into ac­count the (de­fault) in­de­pen­dence of Omega’s choice of num­ber and the Num­ber Lot­tery’s choice of num­ber. Suggest­ing that there are only two pos­si­ble strate­gies for ap­proach­ing this prob­lem was a con­se­quence of my poor word­ing.

• I think this line of rea­son­ing re­lies on the Num­ber Lot­tery’s choice of num­ber be­ing con­di­tional on Omega’s eval­u­a­tion of you as a one-boxer or two-boxer.

What? I can’t even parse that.

There IS a num­ber in the box which is the same as the one at the Lot­tery Bank. The num­ber ei­ther is prime or it is com­pos­ite.

Ac­cord­ing to the hy­po­thet­i­cal, if I two-box, there is a 99.9% cor­re­la­tion with Omega putting a com­pos­ite num­ber in his box, in which case my pay­ooff is \$2,001,000. There is a 0.1% cor­re­la­tion with Omega putting a prime num­ber in the box in which case my pay­offis \$1,001,000. If the cor­re­la­tion is a good es­ti­mate of prob­a­bil­ity, then my ex­pected pay­off from two-box­ing is \$2mil­lion more or less. If I one-box, blah blah blah ex­pected pay­off is \$1mil­lion.

• Sorry for my poor phras­ing. The Num­ber Lot­tery’s num­ber is ran­domly cho­sen and has noth­ing to do with Omega’s pre­dic­tion of you as a two-boxer or one-boxer. It is only Omega’s choice of num­ber that de­pends on whether it be­lieves you are a one-boxer or two-boxer. Does this clear it up?

Note that there is a caveat: if your strat­egy for de­cid­ing to one-box or two-box de­pends on the out­come of the Num­ber Lot­tery, then Omega’s choice of num­ber and the Lot­tery’s choice of num­ber are no longer in­de­pen­dent.

• By that logic, you would not chew gum in the CGTA prob­lem.

• A cru­cial differ­ence be­tween the two prob­lems is that in the CGTA prob­lem I can dis­t­in­guish among an in­cli­na­tion to chew gum, the act of chew­ing gum, the de­ci­sion to chew gum, and the means whereby I reach that de­ci­sion. If the cor­rect causal story is that CGTA causes both ab­scesses and an in­cli­na­tion to chew gum, and that gum-chew­ing is pro­tec­tive against ab­scesses in the en­tire pop­u­la­tion, then the cor­rect de­ci­sion is to chew gum. The pres­ence of the in­cli­na­tion is bad news, but not the de­ci­sion, given knowl­edge of the in­cli­na­tion. This is the causal di­a­gram:

My ac­tion in look­ing at this di­a­gram and mak­ing a de­ci­sion on that ba­sis does not ap­pear in the di­a­gram it­self. If it were added, it would be only as part of an “Other fac­tors” node with an ar­row only to “Ac­tion of chew­ing gum”. This in­tro­duces no prob­le­matic en­tan­gle­ments be­tween my de­ci­sion pro­cesses and the avoidance of ab­scesses. Perfect ra­tio­nal­ity is as­sumed, of course, so that when con­sid­er­ing the health risks or benefits of chew­ing gum, I do not let my de­ci­sion be ir­ra­tionally swayed by the pres­ence or ab­sence of the in­cli­na­tion, only ra­tio­nally swayed by what­ever the ac­tual causal re­la­tion­ships are.

In the pre­sent thought-ex­per­i­ment, it is an es­sen­tial part of the setup that these dis­tinc­tions are not pos­si­ble. My ac­tual de­ci­sion pro­cesses, in­clud­ing those re­sult­ing from con­sid­er­ing the causal struc­ture of the prob­lem, are by hy­poth­e­sis a part of that causal struc­ture. That is what Omega uses to make its pre­dic­tion. This is the causal di­a­gram I get:

No­tice the ar­row from the di­a­gram it­self to my de­ci­sion-mak­ing, which makes a non-stan­dard sort of causal di­a­gram. To ac­tu­ally prove that some strat­egy is cor­rect for this prob­lem re­quires for­mal­is­ing such self-refer­en­tial de­ci­sion prob­lems. I am not aware that any­one has suc­ceeded in do­ing this. For­mal­is­ing such rea­son­ing is part of the FAI prob­lem that MIRI works on.

Hav­ing drawn that causal di­a­gram, I’m now not sure the origi­nal prob­lem is con­sis­tently for­mu­lated. As stated, Omega’s num­ber has no causal en­tan­gle­ment with the Lot­tery’s num­ber. But my de­ci­sion is so en­tan­gled, and quite strongly. For ex­am­ple, if the Lot­tery chooses a num­ber whose prime­ness I can in­stantly judge, I know what I will get from the Lot­tery, and if Omega has cho­sen a differ­ent num­ber whose prime­ness I can­not judge, the prob­lem is then re­duced to plain New­comb and I one-box. If Omega’s de­ci­sion shows such perfect statisi­cal de­pen­dence with mine, and mine is strongly statisi­cally de­pen­dent with the Lot­tery, Omega’s de­ci­sion can­not be statis­ti­cally in­de­pen­dent of the Lot­tery. But by the Markov as­sump­tion, de­pen­dence im­plies causal en­tan­gle­ment. So must that as­sump­tion be dropped in self-refer­en­tial causal de­ci­sion the­ory? Some­thing for MIRI to think about.

The case where I can see at a glance whether Omega’s num­ber is prime is like New­comb with a trans­par­ent box B. I (me, not the hy­po­thet­i­cal par­ti­ci­pant) de­cline to offer a de­ci­sion in that case.

• This is my rea­son­ing ex­actly, as I rea­soned it out, be­fore read­ing com­ments. This seems like the “ob­vi­ous” an­swer, at least in the time­less sense. Of course, when read­ing a prob­lem like this, one more thing you have to bear in mind (out­side the prob­lem, of course, as in­side the prob­lem you do not have this ad­di­tional in­for­ma­tion) is that the ob­vi­ous solu­tion is un­likely to be fully cor­rect, or else what would be the point of Eliezer pos­ing the prob­lem in the first place… If one is go­ing to ac­cept time­less de­ci­sion mak­ing, then I see no rea­son not to ac­cept it in cases where it is an­chored on fun­da­men­tal math­e­mat­i­cal facts rather than “just” on phys­i­cal facts like whether or not you will choose to take the money, as in the stan­dard New­comb’s para­dox.

• This seems like the “ob­vi­ous” an­swer, at least in the time­less sense.

Maybe, but ob­vi­ous an­swers to such prob­lems are of­ten wrong. Often, mul­ti­ple differ­ent an­swers are each ob­vi­ously the ex­clu­sively right an­swer. And look at all the peo­ple in this thread one-box­ing. Not so ob­vi­ous to them.

My rea­son­ing was as stated, but I’m not go­ing to use its “ob­vi­ous­ness” as an ad­di­tional ar­gu­ment in favour of it. And on read­ing the com­ments and the Face­book thread, I no­tice that I have ne­glected to con­sider the hy­po­thet­i­cal situ­a­tions in which the two num­bers are differ­ent. On con­sid­er­ing it, it seems that I should still ar­gue as I did, us­ing all the available in­for­ma­tion, i.e. that on this oc­ca­sion the two num­bers are the same. But it is merely ob­vi­ous to me that this is so; I am not at all cer­tain.

the ob­vi­ous solu­tion is un­likely to be fully cor­rect, or else what would be the point of Eliezer pos­ing the prob­lem in the first place...

I’m dis­in­clined to guess the right an­swer on the ba­sis of pre­dict­ing the hid­den pur­poses of some­one smarter than me. But I can, as it hap­pens, think of a rea­son for pos­ing a ques­tion whose “ob­vi­ous” solu­tion is com­pletely right. It could be just the first of a gar­den path se­ries of puz­zles for which the “ob­vi­ous” solu­tions are col­lec­tively in­con­sis­tent with any known de­ci­sion the­ory.

• But it is merely ob­vi­ous to me that this is so; I am not at all cer­tain.

Upvoted for awe­some epi­gram.

• To prop­erly de­cide I need to know if I am Jones or Leftie.

EDIT: Given that the Ul­ti­mate Trol­ley only di­rectly kills Jones or Lefty, any agent that tries fac­tor­ing ei­ther Omega’s or the lot­tery’s num­ber must be Jones or Leftie. To de­cide whether it’s bet­ter to fac­tor the num­ber and die by trol­ley it’s nec­es­sary to know the an­swer to the Ul­ti­mate Trol­ley Prob­lem and whether it’s bet­ter for Jones or Leftie to die. The ex­pected util­ity of the fi­nal out­come of the Ul­ti­mate Trol­ley Prob­lem cer­tainly dwarfs the ex­pected util­ity of ei­ther the lot­tery or Omega’s bank.

• I would claim that the cor­rect prob­a­bil­ity to hold is some­where around 0.999 in fa­vor of com­pos­ite if you take both boxes [...]

EDIT: Looks like I was right about prob­a­bil­ities, but too hasty about think­ing that meant you should two-box. Omega can be mal­i­cious:

Sup­pose we do this Prime­comb + lot­tery ex­per­i­ment a jillion times. What al­gorithm max­i­mizes pay­out over those jillion times?

One-box­ing sure seems like a good plan—usu­ally the lot­tery will pay out, some­times not, but no big­gie since you can’t af­fect it. And since there aren’t that many prime num­bers, the lot­tery and the box don’t share num­bers very of­ten, though when they do you always lose the lot­tery.

But sup­pose you de­cide to two-box ev­ery time you see the lot­tery and the box have the same num­ber. Now Omega’s ac­tion is un­defined—if the lot­tery num­ber is com­pos­ite Omega can ba­si­cally choose whether you’re to two-box or one-box. If we think in terms of the jillion tri­als of the same game, one-box­ing would still be bet­ter, since when Omega un­definedly de­cides to make you two-box, you were go­ing to win the lot­tery any­how and could have got­ten more money if Omega had de­cided to make you one-box.

How­ever, if you two-box ev­ery time the num­bers are the same, ev­ery time the num­bers are the same you’ll win the lot­tery. So if you see the num­bers the same, it cer­tainly sounds rea­son­able to try to be part of the lot­tery-win­ning group, right?

Hold up though. Sup­pose we get to pro­gram Omega a lit­tle bit. One ver­sion we make nice—call it Nice­mega. It never makes the num­bers be the same. so I always one-box and get lots of money. Another ver­sion we make mean—Mean­mega. It chooses the num­ber on the box to min­i­mize the money it has to pay out. If you two-box when the num­bers are the same, it makes you two-box when­ever it can. If you are will­ing to two-box when the num­bers are the same and you start see­ing a lot of same num­bers, you should switch plans, be­cause you’re prob­a­bly get­ting Mean­mega’d! So why should you two-box when you see the num­bers the same, if it just means you’re get­ting Mean­mega’d?

In other words, the op­ti­mal strat­egy re­ally can be globally op­ti­mal, even though some­times it re­quires you to take lo­cally bad ac­tions. Seem a lit­tle fa­mil­iar?

My fa­vorite ex­am­ple for this is the Un­ex­pected Hang­ing para­dox.

A judge tells a con­demned pris­oner that he will be hanged at noon on one week­day in the fol­low­ing week but that the ex­e­cu­tion will be a sur­prise to the pris­oner. He will not know the day of the hang­ing un­til the ex­e­cu­tioner knocks on his cell door at noon that day.

Hav­ing re­flected on his sen­tence, the pris­oner draws the con­clu­sion that he will es­cape from the hang­ing. His rea­son­ing is in sev­eral parts. He be­gins by con­clud­ing that the “sur­prise hang­ing” can’t be on Fri­day, as if he hasn’t been hanged by Thurs­day, there is only one day left—and so it won’t be a sur­prise if he’s hanged on Fri­day. Since the judge’s sen­tence stipu­lated that the hang­ing would be a sur­prise to him, he con­cludes it can­not oc­cur on Fri­day.

He then rea­sons that the sur­prise hang­ing can­not be on Thurs­day ei­ther, be­cause Fri­day has already been elimi­nated and if he hasn’t been hanged by Wed­nes­day night, the hang­ing must oc­cur on Thurs­day, mak­ing a Thurs­day hang­ing not a sur­prise ei­ther. By similar rea­son­ing he con­cludes that the hang­ing can also not oc­cur on Wed­nes­day, Tues­day or Mon­day. Joyfully he re­tires to his cell con­fi­dent that the hang­ing will not oc­cur at all.

The next week, the ex­e­cu­tioner knocks on the pris­oner’s door at noon on Wed­nes­day — which, de­spite all the above, was an ut­ter sur­prise to him. Every­thing the judge said came true.

The ques­tion is—where was the flaw in the pris­oner’s logic? The an­swer: the only flaw is that the judge re­ally was pre­pared to hang the man on Fri­day, even though by then it’s not a sur­prise. Every pris­oner you hang with­out sur­prise on Fri­day buys you four to hang with gen­uine sur­prise on the other days of the week. If you’re un­will­ing to pay in the coin of failed sur­prises, you can­not buy gen­uine ones. If you’re the judge, you roll a five-sided die and hang the pris­oner on the cor­re­spond­ing day. If you roll Fri­day, then you damn well hang them on Fri­day to no sur­prise, or else you’re not even re­ally try­ing.

A similar logic leads to the re­jec­tion of two-box­ing when you see that the num­bers are the same. If you aren’t will­ing to one-box when the lot­tery has rol­led a Fri­day, er, a prime num­ber, (and Omega has de­cided to be be a jerk and rub it in) then you’re not ever ac­tu­ally one-box­ing.

• This post al­most con­vinced me. I was think­ing about it in terms of a similar al­gorithm, “one-box un­less the num­ber is ob­vi­ously com­pos­ite.” Your ar­gu­ment con­vinced me that you should prob­a­bly one-box even if Omega’s num­ber is, say, six. (Even leav­ing aside the fact that I’d prob­a­bly mess up more than one in a thou­sand ques­tions that easy.) For the rea­sons you said, I ten­ta­tively think that this al­gorithm is not ac­tu­ally one-box­ing and is sub­op­ti­mal.

But the al­gorithm “one-box un­less the num­bers are the same” is differ­ent. If you were play­ing the reg­u­lar New­comb game, and some­one cred­ibly offered you \$2M if you two-box, you’d take it. More to the point, you pre­sum­ably agree that you should take it. If so, you are now op­er­at­ing on an al­gorithm of “one-box un­less some­one offers you more money.”

In this case, it’s just like they are offer­ing you more money: if you two-box, it’s com­pos­ite 99.9% of the time, and you get \$2M.

The one thing we know about Omega is that it picks com­pos­ites iff it pre­dicts you will two-box. In the Mean­mega ex­am­ple, it picks the num­bers so that you two-box when­ever it can, that just means when­ever the lot­tery num­ber is com­pos­ite. So in all those cases, you get \$2M. That you would have got­ten any­way. Huh. And \$1M from one-box­ing if the lot­tery num­ber is prime. Whereas, if you one-box, you get \$1M 99.9% of the time, plus a lot of money from the lot­tery any­way. OK, so you’re com­pletely right. I might have to think about this more.

As­sum­ing Man­fred is com­pletely right, how many non-iden­ti­cal num­bers should it take be­fore you de­cide you’re not deal­ing with Mean­mega and can start two-box­ing when they’re the same?

• Omega’s num­ber is cho­sen in­de­pen­dently of the lot­tery num­ber, how­ever.

That’s im­pos­si­ble (given your other speci­fi­ca­tions) against an agent who two-boxes iff the lot­tery num­ber and Omega’s num­ber match. If the lot­tery num­ber is prime, this causes Omega to en­sure that a differ­ent prime num­ber is placed in its box.

Edit: Real­ized; it should be “The lot­tery num­ber is cho­sen in­de­pen­dently of Omega’s num­ber.”

Edit again: Just re­al­ized that this phrase sneaks in a causal pos­tu­late: Omega can’t change the out­put of the lot­tery! Start­ing here in rea­son­ing might make the prob­lem a lot eas­ier. In “log­i­cal time,” first the lot­tery num­ber is se­lected, then you make your de­ci­sion, then Omega makes eirs. Of course, for­mal­iz­ing this is non-triv­ial.

• I want to pre-com­mit to 1-box­ing as long as the num­bers are differ­ent, just like in the stan­dard New­comb prob­lem. Since Omega knows that I will no­tice if the num­bers are the same, I can de­cide to make a spe­cial case for this situ­a­tion with­out af­fect­ing the stan­dard case. But I still want to pre-com­mit to 1-box­ing in this case, for the same rea­son I want to pre-com­mit in stan­dard New­comb: I can pre­dict that Omega will be much more likely to put \$1,000,000 in box B if I do so, and the causal­ity here doesn’t al­low me to in­fluence the out­come of the lot­tery.

• You see two boxes and you can ei­ther take both boxes, or take only box B. Box A is trans­par­ent and con­tains \$1000. Box B con­tains a visi­ble num­ber, say 1033. The Bank of Omega, which op­er­ates by very clear and trans­par­ent mechanisms, will pay you \$1M if this num­ber is prime, and \$0 if it is com­pos­ite. Omega is known to se­lect prime num­bers for Box B when­ever Omega pre­dicts that you will take only Box B; and con­versely se­lect com­pos­ite num­bers if Omega pre­dicts that you will take both boxes. Omega has pre­vi­ously pre­dicted cor­rectly in 99.9% of cases.

At­tempt­ing to trans­late this English de­scrip­tion into a pro­gram-seg­ment which I can do alge­bra on, I get a type er­ror. I can only re­solve the type er­ror by chang­ing a vi­tal as­pect of the rules, and I have sev­eral op­tions for how to do so and no prior pro­vided, so this ques­tion is unan­swer­able as writ­ten. This is a very com­mon prob­lem with de­ci­sion the­ory work, and I think ev­ery­one should make a habit of writ­ing de­ci­sion the­ory ques­tions as stat­i­cally-typed pro­grams, not as prose.

The is­sue is that, in or­der to pre­dict whether you will take one or both boxes, Omega must sup­ply all the in­puts to your simu­la­tion, in­clud­ing the num­ber that you see; and one of the in­puts, is Omega’s own out­put. Re­plac­ing the num­ber with a boolean that you don’t get to look at would re­solve the is­sue, and you al­most do that, by say­ing that you’re not al­lowed to fac­tor the num­ber, but the prob­lem still fails to com­pile if you en­tan­gle your de­ci­sion with any prop­erty of the num­ber that’s even a lit­tle bit re­lated to prime­ness.

• the prob­lem still fails to com­pile if you en­tan­gle your de­ci­sion with any prop­erty of the num­ber that’s even a lit­tle bit re­lated to primeness

That doesn’t seem com­pletely right to me. For ex­am­ple, odd­ness is re­lated to prime­ness. If I wanted to do the op­po­site of what Omega pre­dicted, I might try to one-box on even num­bers and two-box on odd num­bers. But then Omega can just give me an odd num­ber that isn’t prime. More gen­er­ally, if we drop the lot­tery and sim­plify the prob­lem to just trans­par­ent New­comb’s with prime/​com­pos­ite, then for any player strat­egy that isn’t ex­actly “two-box if prime, one-box if com­pos­ite”, Omega can find a way to be right.

Another prob­lem is that Omega might have mul­ti­ple ways to be right, e.g. if if your strat­egy is “one-box if prime, two-box if com­pos­ite” or “one-box if odd, two-box if even”. But then it seems that re­gard­less of how Omega chooses to break ties, as long as it pre­dicts cor­rectly, one-box­ers can­not lose out to other strate­gies. That ap­plies to the origi­nal prob­lem as well, so I’m in fa­vor of one-box­ing there (see wedrifid’s and Carl’s com­ments for de­tails).

Over­all I agree that giv­ing an un­der­speci­fied prob­lem and pa­per­ing it over with “you don’t have a calcu­la­tor” isn’t very nice, and it would be bet­ter to have well-speci­fied prob­lems in the fu­ture. For ex­am­ple, when Gary was de­scribing the trans­par­ent New­comb’s prob­lem, he was care­ful to say that in the simu­la­tion both boxes are full. In our case the prob­lem turned out to be kinda sorta solv­able in the end, but I guess it was just luck.

• Yep, this all seems cor­rect; the player does not have enough de­grees of free­dom to pre­vent there from be­ing a fix­point, and it is pos­si­ble to prove for all in­ter­pre­ta­tions that no strat­egy does bet­ter than ty­ing with the sim­ple one-box strat­egy. But I feel, very strongly, that al­low­ing this par­tic­u­lar kind of am­bi­guity into de­ci­sion the­ory prob­lems is a re­li­ably los­ing move. That road leads only to con­fu­sion, and that par­tic­u­lar mis­take is re­spon­si­ble for many (pos­si­bly most) pre­vi­ous failures to figure out de­ci­sion the­ory.

• The is­sue is that, in or­der to pre­dict whether you will take one or both boxes, Omega must sup­ply all the in­puts to your simu­la­tion, in­clud­ing the num­ber that you see

There doesn’t need to be a con­crete simu­la­tion where all vari­ables at­tain canon­i­cal val­ues. In­stead, some vari­ables can re­tain their sym­bolic defi­ni­tions, in­clud­ing as re­sults of re­cur­sive calls. Such pro­grams can some­times be eval­u­ated even with­out ex­plic­itly pos­ing the prob­lem of the ex­is­tence of con­sis­tent vari­able as­sign­ments, es­pe­cially if you are al­lowed to make some op­ti­miz­ing trans­for­ma­tions dur­ing the eval­u­a­tion (elimi­nat­ing vari­ables with­out eval­u­at­ing them).

• It’s also more than an un­known boolean (prime or not) be­cause you can check if it’s the same num­ber as the lot­tery out­put.

• On one hand, I sym­pa­thize with your ar­gu­ment. When Gary was de­sign­ing the trans­par­ent New­comb’s prob­lem, he was care­ful to point out that the simu­la­tion sees both boxes as full.

On the other hand, can you point out ex­actly where Carl’s solu­tion pro­posed on the Face­book thread dis­agrees with your claim that the prob­lem is un­solv­able?

• (Note: I haven’t checked yet to see if 1033 is prime)

So… ba­si­cally, it’s the stan­dard New­comb’s prob­lem, one box or two, one box­ing means it’s a prime num­ber and two box­ing means it’s a com­pos­ite num­ber be­ing dis­played for the lot­tery, in this sin­gu­lar case.

I’d still prob­a­bly one box here. If 1033 is prime, and I two box… well, then, Omega prob­a­bly wouldn’t have picked it and we wouldn’t be dis­cussing this sce­nario.

Put an­other way, I don’t see how the lot­tery num­ber match­ing Omega’s num­ber gives me any use­ful in­for­ma­tion about Omega’s ac­cu­racy, since the value of one num­ber in no way de­pends on the other.

• I’ll flip a coin. Heads I 2-box, tails I 1-box. That’s got to be pretty good in ex­pec­ta­tion. Gotta make omega work for that 99.9% ac­cu­racy!

• Prior to see­ing the fact that the Lot­tery num­bers matched, I would have liked to have pre-com­mit­ted to one box­ing in all cases. That’s how I set my de­ter­minis­tic al­gorithm.

There­fore, I will surely one box. This seems more or less iden­ti­cal to the clas­sic New­comb. Yes, I know the num­ber is prime now, so I would like to get away with tak­ing the sec­ond box and I should try as hard as I can to over­ride my ini­tial pro­gram­ming and two box...but un­less my al­gorithm is un­suc­cess­fully pre-com­mit­ted, I will fail to do so.

Some folks seem to think you aught to pre-com­mit to set your al­gorithm to two box if and only if the num­bers match. That’s wrong be­cause you aren’t effect­ing the Lot­tery. All you are do­ing is mak­ing it so that Omega some­times chooses a com­pos­ite num­ber which is iden­ti­cal to the Lot­tery when the Lot­tery is a com­pos­ite num­ber. The Lot­tery is ran­dom and ir­rele­vant.

Edit: Some other com­men­ta­tors seem to in­ter­pret EDT as two box­ing in this sce­nario, so I guess it does differ from clas­si­cal New­comb. Would EDT also re­quire you to pre-com­mit to two-box­ing, or is that just what EDT says when thrust into the sce­nario? (if the lat­ter, isn’t that a huge prob­lem?)

• Since this is too com­pli­cated for me to figure out in any deep sense in 2 min­utes (and I’m not sure ex­actly what com­pu­ta­tion Omega is do­ing or if it’s even well-defined), I’m fal­ling back on EDT, and two-box­ing to get the \$2M lot­tery. EDT might be sub­op­ti­mal in a lot of cases (smoker’s le­sion), but at least it’s un­likely to choose wrong, which is bet­ter than I can say for what I’d pick if I tried to use TDT or CDT-com­bined-with-un­cer­tainty-about-who-I-am (whether I’m a copy simu­lated by omega to de­ter­mine pay­offs for the real copy of me).

• Writ­ten be­fore read­ing com­ments; The an­swer was de­cided within or close to the 2 minute win­dow.

I take both boxes. I am un­cer­tain of three things in this sce­nario: 1)whether the num­ber is prime; 2) whether Omega pre­dicted I would take one box or two; and 3) whether I am the type of agent that will take one box or two. If I take one box, it is highly likely that Omega pre­dicted this cor­rectly, and it is also highly likely that the num­ber is prime. If I take two boxes, it is highly likely that Omega pre­dicted this cor­rectly and that the num­ber is com­pos­ite. I pre­fer the num­ber to be com­pos­ite, there­for I take both boxes on the an­ti­ci­pa­tion that when I do so I will (cor­rectly) be able to up­date to 99.9% prob­a­bil­ity that the num­ber is com­pos­ite.

Think­ing this through ac­tu­ally led me to a bit of in­sight on the origi­nal new­comb’s prob­lem, namely that last part about up­dat­ing my be­liefs based on which ac­tion I choose to take, even when that ac­tion has no causal effects on the sub­ject of my be­liefs. Tak­ing an ac­tion al­lows you to strongly up­date on your be­lief about which ac­tion you would take in that situ­a­tion; in cases where that fact is causally con­nected to oth­ers (in this case Omega’s pre­dic­tion), you can then up­date through those con­nec­tions.

• I think most of the com­menters aren’t get­ting that this is a par­ody. Edit: It turns out I was wrong.

• “Un­like these other highly-con­trived hy­po­thet­i­cal sce­nar­ios we in­vent to test ex­treme cor­ner-cases of our rea­son­ing, this highly-con­trived hy­po­thet­i­cal sce­nario is a par­ody. If you ever find your­self in the oth­ers, you have to take it se­ri­ously, but if you find your­self in this one, you are un­der no such obli­ga­tion.”

• Yes, that’s what I’m say­ing. The other ones are meant to prove a point. This one is just to make you laugh, just like the one it is named af­ter. http://​​www.mind­spring.com/​​~mf­pat­ton/​​Tis­sues.htm

• So if you found your­self in the un­likely sce­nario of a reg­u­lar New­comb’s Prob­lem, you have an an­swer for it; but if you found your­self in the un­likely sce­nario of this prob­lem, you wouldn’t feel obliged to be able to an­swer it?

• Well… the linked Ul­ti­mate Trol­ley prob­lem is a par­ody.I

f this is a par­ody, it’s ev­i­dently in­ter­est­ing enough to think about any­way.

• Can you ex­plain why Eliezer’s mo­tives for writ­ing it should limit what any­one else chooses to do with it?

ETA: Par­ent ed­ited. This now makes less sense as a re­sponse.

• yeah, sorry. I re­al­ized that even though the first sen­tence on its own was a sim­ple true state­ment, it might con­no­tate that I thought that ev­ery­one who was tak­ing it se­ri­ously was be­ing silly when I re­ally just meant to in­no­cently point out some ev­i­dence that al­igned with sum­mer­stay’s opinion that it might be par­ody (or se­ri­ous-but-par­ody-mimick­ing, or some­thing). So I added a sec­ond sen­tence to dis­as­so­ci­ate my­self from the con­no­ta­tion that might oth­er­wise be in­ferred.

• I’m at the cur­rent MIRI work­shop, and the Ul­ti­mate New­comb’s Prob­lem is not a par­ody.

• I don’t get paid on the ba­sis of Omega’s pre­dic­tion given my ac­tion. I get paid on the ba­sis of my ac­tion given Omega’s pre­dic­tion. I at least need to know the base-rate prob­a­bil­ity with which I ac­tu­ally one-box (or two-box), al­though with only two min­utes, I would prob­a­bly need to know the base rate at which Omega pre­dicts that I will one-box. Ac­tu­ally, just get­ting the prob­a­bil­ity for each of P(Ix|Ox) and P(Ix|O~x) would be great.

I also don’t have a mechanism to de­ter­mine if 1033 is prime that is read­ily available to me with­out get­ting hit by a trol­ley (with what prob­a­bil­ity do I get hit by the trol­ley, in­ci­den­tally?), nor do I know the ra­tio of odd-num­bered primes to odd-num­bered com­pos­ites is off-hand.

I don’t quite have enough in­for­ma­tion to solve the prob­lem in any sort of re­spectable fash­ion. So what the heck, I two-box and hope that Omega is right and that the num­ber is com­pos­ite. But if it isn’t, then I cry into my mil­lion dol­lars. (With P(.1): I don’t ex­pect to ac­tu­ally be sad win­ning \$1M, es­pe­cially af­ter hav­ing played sev­eral thou­sand times and pre­sum­ably hav­ing won at least some money in that pe­riod.)

• +EV of one box­ing swamps my un­cer­tainty over the sta­tus of 1033.

• Prob­lem lacks speci­fi­ca­tion: ought we as­sume that Omega also pre­dicted TNL’s num­ber? Or was that ran­dom both to me, Omega-past, and TNL? Omega pre­dict­ing cor­rectly in 99.9% of pre­vi­ous cases doesn’t de­ter­mine this.

(eta: Ah, was an­swered on the Face­book thread; Omega pre­dicted the lot­tery num­ber. Hm.)

• 8 Aug 2015 17:29 UTC
0 points

...so if I google ‘is 2033 a prime num­ber’ and re­ceive the an­swer that it isn’t, all in un­der two min­utes, and put the money from box A into box B and then choose box B, do I get any money?:)

• 20 Oct 2013 14:14 UTC
0 points

EDIT:

Read­ing through com­ments it ap­pears the idea is that you’re at­tempt­ing to con­trol whether or not omega matches the num­ber with your choice of strat­egy, rather than what you get from matched or un­matched lot­ter­ies. So the idea seems to be that always tak­ing onebox yields lower pay­offs from match­ing lot­ter­ies but causes lot­ter­ies to match less of­ten, which is benefi­cial be­cause un­match­ing lot­ter­ies have bet­ter pay­offs than match­ing lot­ter­ies.

• Numer­i­cal Lot­tery has ran­domly se­lected 1033...

… if you try to fac­tor the num­ber you will be run over by the trol­ley from the Ul­ti­mate Trol­ley Prob­lem.

This game seems to have a roughly 50% chance of a fatal­ity.

• I haven’t looked at the com­ments yet. Two min­utes was enough think­ing time to get me to one-box, put not enough time to ver­bal­ize my in­tu­ition. There was a post ear­lier on lw about mak­ing de­ci­sions with im­perfect mem­ory that is rele­vant, I think.

My in­tu­ition is some­thing along the lines of ‘my de­ci­sions only af­fect whether omega gives me a mil­lion dol­lars or not, so the lot­tery doesn’t mat­ter’.

Fur­ther thoughts: your strat­egy when num­bers match change the in­for­ma­tion that num­bers match­ing con­veys. If you one box, it tells you that the par­tic­u­lar round loses the lot­tery. If you two box, it tells you that the par­tic­u­lar round wins.

Since you win the lot­tery just as of­ten re­gard­less of when you know that you win the lot­tery, the value of this in­for­ma­tion is zero. All two-box­ing does is move the co­in­ci­dences to co­in­cide with win­ning the lot­tery. That, and lose you nearly a mil­lion dol­lars per co­in­ci­dence.

• (An­swer­ing be­fore read­ing any other re­sponses)

Both boxes—I want the num­ber to be com­pos­ite, so I want Omega to have se­lected a com­pos­ite num­ber, which he’d have more chances of do­ing if I two-boxed.

EDIT: wedrifid’s ex­pla­na­tion has now mostly con­vinced me that one-box­ing is cor­rect in­stead. (My ex­pressed logic was too much EDT-in­fluenced, I think)

• Both boxes—I want the num­ber to be com­pos­ite, so I want Omega to have se­lected a com­pos­ite num­ber, which he’d have more chances of do­ing if I two-boxed.

You want the num­ber to be com­pos­ite, so you also want the Numer­i­cal Lot­tery’s ran­dom num­ber gen­er­a­tor to have se­lected a com­pos­ite num­ber. That’s trick­ier to in­fluence.

• The num­ber 1033 is prime. The rest of the hy­po­thet­i­cal sce­nario is pretty con­fus­ing, and it’ll take me longer to an­a­lyze it and be con­fi­dent about my con­clu­sion than it did to de­ter­mine that 1033 is prime.

• The num­ber 1033 is prime. The rest of the hy­po­thet­i­cal sce­nario is pretty con­fus­ing, and it’ll take me longer to an­a­lyze it and be con­fi­dent about my con­clu­sion than it did to de­ter­mine that 1033 is prime.

You just got run over by a trol­ley. Death is al­most cer­tainly worse than ei­ther be­ing given a free \$1M or \$2M. This illus­trates that failure to con­sider the de­ci­sion be­ing made is in gen­eral a sub-op­ti­mal game-the­o­retic strat­egy. For­tu­nately Omega pays out de­spite your failure to take any boxes (be­fore your demise) and so your heirs gain an ad­di­tional \$1M. As for your time­less in­fluence on Omega’s de­ci­sion: by the word­ing given (“IF … AND IF” in­stead of “IF … ELSE”) Omega is free to choose whichever num­ber he likes given that nei­ther crite­ria is satis­fied.

Rest in Peace.

• 12 Sep 2013 9:45 UTC
0 points

You have two min­utes to make a de­ci­sion, you don’t have a calcu­la­tor, and if you try to fac­tor the num­ber you will be run over by the trol­ley from the Ul­ti­mate Trol­ley Prob­lem.

This isn’t a fair prob­lem (in the sense that you defined some­where else): two peo­ple both choos­ing to take both boxes, but via differ­ent al­gorithms (only one of whom fac­tor­ing the num­ber), will get differ­ent re­wards.

• I don’t know what “try­ing to fac­tor” even would be for a num­ber so small. It just looks like a prime. I may have seen it on a prime num­ber list, or as a prime fac­tor of some­thing, or who knows where. There’s easy to con­struct rules for de­ter­min­ing di­visi­bil­ity by it’s po­ten­tial fac­tors.

One could also use Miller-Rabin pri­mar­ity test, which I in fact hap­pen to have im­ple­mented be­fore. Much of the pub­lic key cryp­tog­ra­phy de­pends on how test­ing a prime is eas­ier than fac­tor­ing a num­ber. I’m pretty sure there is no gen­eral al­gorithm for de­ter­min­ing when­ever an al­gorithm is a good pri­mar­ity test.

(I pre­sume the point is that you aren’t try­ing to de­ter­mine when­ever it is prime or not, which breaks all sorts of as­sump­tions in­her­ent in util­ity max­i­miza­tion)

• If that both­ers you, how about in­stead of dis­play­ing the num­ber, in­stead what you see is the num­ber en­crypted us­ing a key known only to the lot­tery-run­ners and Omega?

• It could be more in­ter­est­ing, though, if it was 7. That may bet­ter demon­strate the in­con­sis­ten­cies in math­e­mat­ics that re­sult from an in­cor­rect hy­po­thet­i­cal about your choice.

In a trans­par­ent New­comb’s, I can sim­ply take one empty box and leave the other, if one box is empty, and take both boxes if they are both full.

• I’m not clear on what con­sti­tutes try­ing to fac­tor the num­ber. It would seem that notic­ing if it was odd wouldn’t count as try­ing to fac­tor it, but what about forms of in­duc­tive rea­son­ing or non-ex­haus­tive heuris­tics?

• 11 Sep 2013 12:16 UTC
0 points

You pre­vi­ously played the game with Omega and the Numer­i­cal Lot­tery a few thou­sand times before

Then there are a cou­ple billion dol­lars in my bank ac­count, and the marginal util­ity of one more mil­lion wouldn’t be that large. :-)

• That’s not a bug, it’s a fea­ture. If you’re already a billion­aire, then your util­ity func­tion is nice and lin­ear for changes on the or­der of a mil­lion or two, so no need for angst about \$2M not be­ing close to twice as good as \$1M.

• I hadn’t con­sid­ered this an­gle, but I agree with this. If I’m go­ing to get trol­lied for think­ing the wrong thought, I would want to try hard not to think at all since I might ac­ci­den­tally think about how I already know whether the num­ber is prime.

On the other hand, I don’t think this is a con­se­quence the post in­tended.

• Omega knows I’m smart enough to two box when I see that the #’s match up. So the # will be com­pos­ite, thereby fulfilling his goal of pre­dict­ing my ac­tions and re­spond­ing ap­pro­pri­ately. The Lot­tery Bank doesn’t care about my de­ci­sions.

So I’ll two box.. I’ll do so, and would pre­dictably do so. So Omega has se­lected a com­pos­ite #, and I re­ceive 1000\$ from the boxes. Since its the same # the lot­tery will give me 2 mil­lion dol­lars.

• But in this ex­am­ple Omega has already se­lected 1033, a prime, which means Omega already knows you will one-box. If it were me, how I would switch from your op­ti­mal strat­egy to de­ter­minis­ti­cally one-box­ing be­fore even know­ing whether the num­ber is a prime (I didn’t know un­til I looked up; I’m speak­ing from the morgue right now) is be­yond me.

I guess the les­son in this ex­er­cise is the same gen­eral ar­gu­ment Eliezer has made against lot­ter­ies: don’t bet on out­comes you don’t con­trol.

• Im­me­di­ate thoughts, be­fore read­ing com­ments: One-box. I had started to think more deeply un­til I read the part about be­ing run over for fac­tor­ing, and for some rea­son my brain ap­plied it to rea­son­ing about this topic as a whole and spit out a fi­nal an­swer.

In­tu­itively, it seemed one box­ing would get me a mil­lion, as per stan­dard New­comb. The lot­tery two mil­lion seemed like gravy above that (diminish­ing marginal util­ity of money), with a po­ten­tial for 3 mil­lion to­tal. Since they’re in­de­pen­dent, the word “sep­a­rately” and its de­scrip­tion made it seem like the lot­tery was un­able to be af­fected by my ac­tions at all. Thus, take box B, and hope for a lot­tery win. Definitely don’t over think it, or risk a trol­ley en­counter.

• 10 Sep 2013 9:18 UTC
0 points

Post­ing be­fore check­ing the com­ments.

If I take only box B I will ei­ther make 1M\$ or 2M\$. Omega, with its 99,9% ac­cu­racy, will likely have se­lected a prime num­ber. Ex­pected util­ity is 0.999 1M + 0.001 2M \$ = 1M+1K \$.

If I take both, I will ei­ther get 1M+1K \$ or 2M +1K \$. Already I’m grab­bing both boxes, be­cause the ex­pected util­ity is clearly higher. Omega would likely have se­lected a com­pos­ite num­ber. Ex­pected util­ity is there­fore 0.999 2001 K + 0.001 1001 K = 2M \$.

In cases where the Lot­tery num­ber doesn’t match Omega’s, I have a num­ber of gen­eral strate­gies available, most of which might get me hit by the trol­ley de­pend­ing on how it defines fac­tor­ing. Does check­ing whether the ones digit is even or the num­ber 5 count? Does sum­ming the digits (and then the digits of the sum re­cur­sively if needed) and check­ing if the re­sult is 3, 6 or 9 count? Us­ing these two strate­gies would im­prove my odds sig­nifi­cantly, but risks the wrath of the trol­ley.

• 2-box on one-off [Edit: on reread­ing, this comes off as more con­fi­dent than I in­tended. This was what I thought I would do in the 2 min­utes, which in ret­ro­spect were spent un­pro­duc­tively], but the one-off na­ture of the prob­lem is mod­ified by the third para­graph, which means that strat­egy might mean one-box­ing pre­vi­ously got me no money. I would in­ter­pret that as less-than-perfect ac­cu­racy (which might be the source of the 99.9% prob­a­bil­ity) How does Omega deal with mixed strate­gies?

• How does Omega deal with mixed strate­gies?

In nor­mal New­comb, I be­lieve the stan­dard treat­ment is that e leaves the black box empty in those cases. So, in this prob­lem, I guess e would un­con­di­tion­ally se­lect a com­pos­ite num­ber for eir box. With that speci­fi­ca­tion, (two-box­ing un­con­di­tion­ally) weakly dom­i­nates any un­con­di­tional mixed strat­egy, both in origi­nal New­comb and this prob­lem.

• I did not in­ter­pret para­graph 3 to con­tain any in­for­ma­tion about prior pay­outs… For in­stance, if one were to 1-box (suc­cess­fully!) in ev­ery case that did not have such a lot­tery hedge, it would ap­pear con­sis­tent with the prob­lem state­ment to me.

• You pre­vi­ously played the game with Omega and the Numer­i­cal Lot­tery a few thou­sand times be­fore.

it tells you that there were prior pay­outs.

• Ah, well, it tells us that there were prior games.

• If I didn’t get prior pay­outs from those games, the up­dates on that is way big­ger than any other rea­son­ing such as what we’re do­ing on this thread.

• I ex­pect to main­tain con­trol over Omega(to­day, my de­ci­sion), even af­ter learn­ing Omega(to­day, my de­ci­sion) = Lot­tery(to­day). I take both boxes for ~\$2,001,000.

Since I don’ t know the defi­ni­tion of the pro­cess gen­er­at­ing the lot­tery num­ber, my prior dis­tri­bu­tion over the pri­mal­ity of to­day’s lot­tery num­ber (be­fore learn­ing that Omega’s num­ber in Box B and the Lot­tery Num­ber are equal) would be taken from the dis­tri­bu­tion of primes (1/​log(1033) ~ 0.332). I don’t know why my in­tu­ition says to throw that num­ber out af­ter I con­di­tion on the equal­ity. I feel less cer­tain that I should throw away my prior on the pri­mal­ity of the lot­tery num­ber if that prior is higher. If I start out with P(prime?(Lot­tery())) =1, and then throw that away be­cause I think I main­tain con­trol of the num­ber’s pri­mal­ity, that seems al­most as in­co­her­ent as if I gave differ­ent dis­tri­bu­tions for the pri­mal­ity of equal num­bers gen­er­ated two differ­ent ways.

This is not a well formed prob­lem. If my strat­egy is to 1-box iff the num­bers match, then omega, choos­ing his num­ber in­de­pen­dently of the lot­tery num­ber must choose a com­pos­ite num­ber, since I will 2-box 99.9% of the time. There­fore if omega is cor­rect 99.9% of the time when the num­bers match, then the lot­tery num­ber must be com­pos­ite at least 99.9% of the time.

How­ever, if my strat­egy is to 2-box iff the num­bers match, then omega, choos­ing his num­ber in­de­pen­dently of the lot­tery num­ber must choose a prime num­ber, since I will 1-box 99.9% of the time. There­fore if omega is cor­rect 99.9% of the time when the num­bers match, then the lot­tery num­ber must be prime at least 99.9% of the time.

I don’t know the odds of the lot­tery, but it can­not be both prime at least 99.9% of the time and com­pos­ite at least 99.9%, so one of these two strate­gies will con­sis­tently make omega when the boxes match.

• I re­al­ized this is slightly wrong as writ­ten, be­cause Omega doesn’t have to be cor­rect 100% of the time when they don’t match, so he could do a lit­tle bet­ter us­ing a ran­dom al­gorithm, but this just means that some of the 99.9%s need to be re­placed with 99.8%.

• I don’t think I un­der­stand the prob­lem. While read­ing the Numer­i­cal Lot­tery (NL) para­graph, I de­cided choos­ing only box B was the ob­vi­ous an­swer, which led me to think I’ve mi­s­un­der­stood some­thing.

In box B is \$1,000, a com­pos­ite num­ber. If I pick a com­pos­ite num­ber, the NL gifts me \$2 mil­lion it-doesn’t-re­ally-mat­ter-what-currency

Oh, I mis­read the prob­lem. In box A is \$1,000. Well, now I think I’ve un­der­stood the prob­lem, and cho­sen the right an­swer for mis­taken rea­sons.

If the NL pays “[me] \$2 mil­lion if it has se­lected a com­pos­ite num­ber, and oth­er­wise [...] \$0,”* and the num­ber it has se­lected goes in box B, then re­gard­less of the num­ber shown I can only profit from the NL by choos­ing box B. I’m guaran­teed a profit by sig­nal­ling to Omega that I will only choose box B.

Even in a sce­nario where box B con­tains the lesser amount, ‘tis still the most ra­tio­nal choice, con­sid­er­ing I can ap­par­ently play the game an in­finite num­ber of times—or at least three thou­sand and one times. Con­sid­er­ing that I will always choose box B when rea­son­ing from the pro­vided in­for­ma­tion (un­less I’m still not un­der­stand­ing some­thing), by this point I have at least \$3,001,000,000 dol­lars; if I ever choose oth­er­wise, Omega will no longer have rea­son to pre­dict I will only one-box, and I lose my guaran­tee. The word ‘Ul­ti­mate’ makes me think I’m dras­ti­cally wrong.